The integral equation with a singularity is as follows: $$\Omega=\int^\pi_0\frac{\cos(\theta)}{\sqrt{1-\cos(\theta)}}\,d\theta$$ I want to calculate this integral numerically. Is there an analytic expression that can be used when $\theta=0$, or is there a way to remove the singularity? Any insights on dealing with singular numerical integrals are very useful to me.
Thank you
On
If you look at the expansion of the integrand around $\theta=0$, you have $$\frac{\cos(\theta)}{\sqrt{1-\cos(\theta)}}=\frac{\sqrt{2}}{\theta }-\frac{11}{12 \sqrt{2}} \theta +O\left(\theta ^3\right)$$ which shows how you can remove the singularity.
$$\int_0^\pi \left(\frac{\cos(\theta)}{\sqrt{1-\cos(\theta)}} -\frac{\sqrt{2}}{\theta }\right)\, d\theta=-\sqrt{2} (2+\log (\pi )-2\log (2))$$
Edit
$$\Omega(\epsilon)=\int^\pi_\epsilon\frac{\cos(\theta)}{\sqrt{1-\cos(\theta)}}\,d\theta=2\frac{ \sin \left(\frac{\epsilon }{2}\right) \log \left(\cot \left(\frac{\epsilon }{4}\right)\right)- \sin (\epsilon )}{\sqrt{1-\cos (\epsilon )}}$$ If you let $\epsilon=4 \tan ^{-1}(t)$ $$\Omega(t)=\sqrt{2} \left(2-\frac{4}{t^2+1}-\log (t)\right)$$
If $\theta$ is close to $0$, then $\cos \theta$ behaves like $1-\frac{\theta^2}{2}$. It means that the integrand behaves like $\frac{2-\theta^2}{\sqrt 2 \theta}$, hence the singularity in zero is non-integrable.